A decomposition of functions with zero means on circles

Ark. Mat., 43 (2005), 383-393

(~) 2005 by Institut Mittag-Leffler. All rights reserved

A d e c o m p o s i t i o n of f u n c t i o n s w i t h zero m e a n s on circles

J o s i p G l o b e v n i k

A b s t r a c t . It is well known that every H61der continuous function on the unit circle is the sum of two functions such that one of these functions extends holomorphically into the unit disc and the other extends holomorphically into the complement of the unit disc. We prove that an analogue of this holds for H61der continuous functions on an annulus A which have zero averages on all circles contained in A which surround the hole.

1. I n t r o d u c t i o n a n d t h e m a i n r e s u l t s

G i v e n a E C a n d CO>0 w r i t e A ( a , C O ) = { ~ c C : ] I - o I < C O } a n d A = A ( 0 , 1). D e n o t e b y B t h e o p e n u n i t b a l l in C 2. A f u n c t i o n f o n a set K c C is c a l l e d H61der c o n t i n u o u s on K ( w i t h e x p o n e n t ~ ) if t h e r e a r e c o n s t a n t s M < : x : a n d c~, 0 < c ~ < l , such t h a t I I ( z ) - f ( w ) l < _ M I z - w l z, w e K .

L e t a E C , 0 > 0 a n d let f b e a H61der c o n t i n u o u s f u n c t i o n f on bA(a, O). It is well k n o w n t h a t

(1.1) f = f+ + f - ,

w h e r e f + a n d f - a r e H S l d e r c o n t i n u o u s f u n c t i o n s on bA(a. P) such t h a t f + (1.2) h a s a c o n t i n u o u s e x t e n s i o n t o ~(a, co) which is h o l o m o r p h i c on A ( a . co);

a n d f

(1.3)

h a s a c o n t i n u o u s e x t e n s i o n t o [C U { ~c }] \ A (a, 0) which is h o l o m o r p h i c on [ C U { o c } ] \ / ~ ( a , CO) a n d v a n i s h e s a t ~c:

a n d t h i s d e c o m p o s i t i o n is unique. In fact,

1 fb f(~)d~ {f+(z), zEA(a,o),

(1.4) 27ri A(a,o) ~ - z - i f ( z ) , z e [ C u { ~ c } l \ A ( a , co).

384 Josip Globevnik

In the present paper we consider functions on the annulus A = { ~ E C : r l < I~l-<r2},

where

O<rl<r2<oc

and ask for a decomposition similar to (1.1). Suppose that f is a HSlder continuous function on the annulus A. For every circle

bA(a, Q)cA

surrounding the origin we have

flbA(a,e)=f+~,e+f~e,

where f~Q satisfies (1.2) and f~,e satisfies (1.3). In general there are no functions f+ and f - on A such t h a t

f+ Ib~(a,e) = f ~

and f-]bZx(~,~) = f~-.~ whenever bA(a, 6)

c A

surrounds the origin. In the present paper we prove that there are such functions f + and f - on A whenever f satisfies

1 f02"

(1.5)

- ~ f(a+ee i~ dO = 0

for every

bA(a, p ) c A

which surrounds the origin:

T h e o r e m 1.1.

Let f be a H61der continuous function on A which satisfies

(1.5)

whenever bA(a,

6 ) c A

surrounds the origin. Then f = f + + f - , where f+ and f - are HSlder continuous functions on A such that for each bA(a,

6 ) c A

which surrounds the origin, f+lbA(a,v) satisfies

(1.2)

and f-lbA(a,e) satisfies

(1.3).

We will also show that, as in the case of the circle, we can view f + and f - as the boundary values of functions, holomorphic on appropriate domains. To describe this, we first rewrite the circle case in a form suitable for generalization.

Let f be a continuous function on

bA(a, p).

The idea is to define a new function F on {(~, ~):~EbA(a, 6)}, that is, on

bA(a, 6)

"lifted" to

~={(~,~):~cc}, by

F(~, ~) = f(~), ~ C bA(a, 6),

[G2] and then to write F as the sum of boundary values of holomorphic functions.

Let

Aa,e

= {(z, w) 9 C2:

(z-a)(w-?t)

= 62}.

The intersection Aa,eNE is the circle {((, ~):ff 9 6)} whose complement in A~,e has two components, A~+e and AS, e, where

Aa, e - { ( z , w ) : ( z - a ) ( w - a ) 62 a n d 0 < l z - a l < 6 } , ha, e = { ( z , w ) : ( z - a ) ( w - ~ ) =62 and 6 <

[z-al}

= {(z,w):

(z-a)(w-gt) =62

and 0 <

Iw-al < 6}

= {(z, w): (~, 2) 9 A~*,e}.

A d e c o m p o s i t i o n o f f u n c t i o n s w i t h z e r o m e a n s o n circles 3 8 5

The sets A+,o and A~,e are closed one-dimensional complex submanifolds of C 2 \ E attached to E along

bA+~ =bAg, o = {(~, ~): ~ C bA(a, Q)}.

It is easy to see that a continuous function h on bA(a, O) satisfies (1.2) if and only if the function H defined on bA+o=bAa, o by H(~, ~) =h(~), ~EbA(a, 0), has a bounded

+ +

continuous extension from bA+,o to Aa,eubAa, ~ which is holomorphic on A+0 [G2].

Similarly, h satisfies (1.3) if and only if H has a bounded continuous extension from bAa, o to A~,oUbA[~,o which is holomorphic on Ag, o and vanishes at oc.

We now pass to functions on A which we will view as functions on A = {(r : r

The set A C E will be a common part of the boundaries of two domains f~+(A) and ~ - ( A ) which we now describe. Let f~+(A) be the union of all Aa+0 such that bA(a, O)CInt A surrounds the origin. Similarly, let ~ - ( A ) be the union of all A~, 0 such that bA(a,o)CIntA surrounds the origin. Clearly, f~-(A) is the image of

~+(A) under the reflection (z, w)~-~(W, 5). It turns out that fl+(A) and ~ - ( A ) are disjoint domains in C 2 \ E attached to E along .~. For each ~ E I n t A there are a neighbourhood UCE of (~, ~) and a wedge with the edge U which is contained in

~+(A). An analogous statement holds for ~ - ( A ) .

T h e o r e m 1.2. Let f be a HSlder continuous function on A which satisfies

(1.5) whenever bA(a, ~ ) c A surrounds the origin. There are a bounded continuous function G + on ~+(A)UbfP(A) which is holoraorphic on ~+(A) and a bounded continuous function G- on ~ - ( A ) U b ~ - ( A ) which is holomorphic on ~ - ( A ) such that

f ( z ) = l G + ( z , 2 ) + l G - ( z , 2 ) , z E A .

z z

Thus, on A the function F(z, 2 ) = f ( z ) is the sum of the boundary values of the holomorphic functions (1/w)G+(z,w) and (1/z)G-(z, w).

2. F o u r i e r c o e f f i c i e n t s o f f u n c t i o n s w i t h z e r o m e a n s Suppose that f is a continuous function on A. For each r, rl <r<_r2, let

i f

ck(r) = ~ e-ik~ i~ dO, k E Z,

7 r

so that ~k~=_~ ck(r)e ik~ is the Fourier series of the function e~~176

We shall need the following description of the Fourier coefficients of functions with zero means on circles.

386 J o s i p G l o b e v n i k

T h e o r e m 2.1. ([G1], [EK], [V])

A continuous function f on A satisfies

(1.5)

for each bA(a,

0 ) c A

surrounding the origin if and only if

(a) c 0 ( r ) = 0 , rl___r_<r2,

(b)

for each nEZ, n~O, there are numbers a,.o, an.a ....

,a,~.lnl_ 1

such that Cn(r)=r-lnl(an,o+anAr2 +...+amlnl_lr2(Inl-1)). ri <_r<_r2.

In the rest of this section we assume that f is a continuous fimction on A which satisfies (1.5) for each

bA(a, Q)cA

surrounding the origin.

If n > l then writing

z=re i~

we get

"

= (a,~,or -2~

+ a n . l r - 2 ( ' - l ) +...

+an,~_lr-2)z ~

= ( a n , o + a,.a a . . . - 1 )

\ z n 5 n z n - - 1 5 n - 1 ~''""~- Z Z z n

1 1

= - . . . + a n n - 1 z n - 1

5

= : P , - l ( z , 1

1/5).

Z

where Pn-1 is a homogeneous polynomial of degree n - 1 . If n < _ - i then we get

Cn

(r)e in~

---- (a~,0 + a , , 1 r 2 +... + a . . . . 1 r 2 ( - n - 1) ) Z - Inl

1/" 1 1

= z Lan'~ ~ 2 S - l + a ' ' l zi-N~_~_2 z +... +a~.n-15 ~-l )

= -Qlnl_l(5,

1 l / z ) ,

Z

where QI,~I-1 is a homogeneous polynomial of degree I n l - 1. Thus, putting

z=re i~

into the series

1 ~ pn(z. 1/5)+l ~ Qn(5.1/z)

(2.1) z z

n=O n=O

we get the Fourier series of the function

ei~176 rl <_r<_r2.

For each t, O < t < l , define the functions ft + and ft- on A as follows

~c 1 cc

(2.2)

f+(rei~ = E tkck(r)eikO = z E tJ+lPJ(Z'

1/5)

k = l j = 0

(2.3)

ft(rei~ = E tlklck(r)eikO = 1 tJ+lOj(5, I/z),

k = - - 3 c Z j=O

A d e c o m p o s i t i o n of functions w i t h zero m e a n s on circles 387

where

z=reWEA

and let

ft(z)=f~-(z)+ft(z), z E A .

Note t h a t for each t, 0 < t < 1, both series above converge uniformly on A. Write

1 / f(~) d~

9 r(z) = ~ / ~(~.o) r r l _ ~ r ~ r 2 ,

and observe that

(2.4)

{ r176176 rl <_r~_r2,

0 < t < l , 0ER,

~ ( ( 1 / t ) r e W ) = - f t ( r e W ) , rl<_r<_r2,

0 < t < l , 0 E R .

3. P r o o f o f T h e o r e m 1.1

Let f be a Hhlder continuous function on A which satisfies (1.5) for each

hA(a, Q)

which surrounds the origin. Then there are P , and Qn as above such that (2.1) with

z=re i~

is the Fourier series of

ei~176 rl<r<<_r2.

Using the decomposition (1.1) on each circle

bA(O,r), rl<r<r2,

we can write

f = f + + f - ,

where for each

r, rl<_r<r2, f+lbzx(o,r)

satisfies (1.2) and f-lbA(0,r) satisfies (1.3).

In fact, the functions f+lbz~(0,~) and -f-Ibz~(0.~) are the limiting values of 4p~(z) as

Izl/*r

and

Izl".xr,

respectively. Since f is Hhlder continuous on A it follows that f + and f - are Hhlder continuous on A [ M , Sections 19 and 20].

For each

r, rl<r<r2,

the function ~ is the Cauchy integral and hence the Poisson integral of

f+lbA(O,r)

SO (2.4) implies that for each

r, r]<r<r2, f~-(re i~

converges uniformly in 0 to

f+ (re i~ as t/~l,

and since f is uniformly continuous on

A,

the standard proof of the boundary continuity of the Poisson integral shows that the convergence is uniform also in

r, rl <r<_r2.

So ft converges to f + uniformly on A as t / ~ l . Similarly ft- converges to f - uniformly on A as t / ' l .

Observe that for each bA(a, Lo) that surrounds the origin, the restriction of 1/~ to

bA(a, O)

satisfies (1.2) and the restriction of 5 to

bA(a,

c0) has a continuous extension to [ C U { o c } ] \ A which is holomorphic on [CU{~c}]\/~. The uniform convergence of the series (2.2) implies that for each t, 0 < t < l ,

f~-IbA(a,o)

satisfies (1.2) whenever

hA(a, ~) c A

surrounds the origin. Similarly, the uniform convergence of the series (2.3) and the multiplication with

1/z

imply that for each t, 0 < t < l ,

f t

]b~(a,o) satisfies (1.3) whenever

bA(a, Q)cA

surrounds the origin. The uniform convergence of ft + to f + and f / to f - as t / ~ l imply that the analogous statements hold for f + and f . This completes the proof.

388 J o s i p G l o b e v n i k

Remark 1. Note that the proof of Theorem 1.1 becomes simpler in the special case when f is smooth, say of class r on A. Recall that putting z=re ie into the series (2.1) we get the Fourier series of the function eie~-4f(re~6). Integrating by parts we see that there is a constant M < o c such that each of the series in (2.1) is dominated on A by the series ~ = 1 M n - 2 which implies that one can define

oo 1

Qn(5,1/z), z c A , f+ (z) = zl ~ Pn(z, 1/5), f - ( z ) = -z

rt~O rt~O

where each of the series converges uniformly on A.

Remark 2. If f in Theorem 1.1 is HSlder continuous on A with exponent a, 0 , = a < l , then for any ~, 0 < / ~ < a , the functions f+ and f - are HSlder continuous on A with exponent ~. This follows from [M, Sections 19 and 20].

4. D o m a i n s ~ + ( A ) a n d ~ - ( A )

We list some simple facts about the domain ~+ (A). Analogous statements hold for ~ - ( A ) , the image of ~+(A) under the reflection (z, w)~-~(~, 5). The proofs are elementary, they can be found in [G2].

Recall that ~+ (A) is defined as the union of all A~+e such that bA(a, 0)CInt A surrounds the origin.

P r o p o s i t i o n 4.1. Let ~f=~(rl+r2). The set ~+(A) is a disjoint union of all 1

A+a,.y such that bA(a, ~)CInt A: it is an unbounded open connected set whose bound- ary consists of ft together with the union of all those A+~,~ for which bA(a, ~f)C A is tangent to both bA(O, rl) and bA(0, r2). For each ~Elnt A there are a neighbourhood U c E of (~, ~), an open cone V in iE, a real two-plane perpendicular to E, and a 5 > 0 such that U + ( V A S B ) c ~ + ( A ) .

P r o p o s i t i o n 4 . 2 . If A a , ~ A 5 . 5 + § then the sets A+~,~ and Ab+~ intersect if and only if a~b and one of the circles bA(a, Q), bA(b, 5) surrounds the other. The sets A+,e and Ag,~ intersect if and only if/~(a, O)N~(b. 5)=0.

Proposition 4.2 implies that f~+(A)Nt2-(A)=O.

5. P r o o f o f T h e o r e m 1.2

Recall that f + is the uniform limit of It + as t / ~ l . where for each t, 0 < t < l , f?(z) = zl ~-~tj+lpj(z, 1/2). z E A ,

j=0

A d e c o m p o s i t i o n o f f u n c t i o n s w i t h z e r o m e a n s o n c i r c l e s 3 8 9

with the series converging uniformly on A. It follows t h a t on

A, 2f+(z)

is a uniform limit of a sequence of polynomials in z and

1/2,

(5.1)

5f+(z)

= lim

Sm(Z,

1/5), Z E A.

rn---~or

Now we reason as in [G2]: The functions

Sm(z, 1/w)

are bounded and continuous

+ +

on Aa,eUbAa, e and holomorphic on A~+,e whenever

bA(a, Q)cA

surrounds the ori- gin. Since A+,e is biholomorphically equivalent to the punctured disc the m a x i m u m principle implies t h a t for each (z, w)EA+,e we have

ISm(z, 1/w)-Sj(z, 1/w)l <

max{iSm (~, 1 / 7 / ) - S j ( ~ , 1/~)1 : (~, 77) 9 bA+,~}

_< m a x { I S m ( ( , 1 / ~ ) - S j ( ~ , 1/~)1 : r 9 A}.

It follows t h a t the sequence

Sin(z, i/w)

converges uniformly on

Q+(A)LJbf~+(A)

to a function

G+(z, w).

Since each

Sin(z, 1/w)

is bounded and continuous on Q+(A)U

bf~+(A)

and holomorphic on 12+(A) the same is true for G +. Obviously, f + ( z ) =

(1/2)G+(z, 5), z 9

In the same way we prove t h a t

f-(z)=(1/z)G-(z,~), z 9

A, where G - is bounded and continuous on

~-(A)UbQ-(A)

and holomorphic on

~ - ( A ) . This completes the proof.

6. D r o p p i n g t h e a s s u m p t i o n on H61der c o n t i n u i t y

T h e m a p ~ + ~ * = 1/~ is the antiholomorphic reflection across bA which fixes bA.

Similarly, given a C C and ~) > 0, the m a p ~ ~-~ ~* -- a + Q2 / ( ~ _ ~) is the antiholomorphic reflection across

bA(a, ~)

which fixes

bA(a, ~)

and m a p s a point on a ray e m a n a t i n g from a at a distance 7 > 0 from a to the point on the same ray at a distance Q2/7 from a.

If f is a continuous function on

bA(a, Q)

which is not necessarily HSlder con- tinuous then the functions f + and f - defined by (1.5) are well defined away from

bA(a, p)

but need not have b o u n d a r y values as we approach

bA(a, Q).

T h e following still holds and is well known:

L e m m a 6.1. ([Z, Vol. 1, p. 288])

Let f be a continuous function on bA(a, Q).

Define f+ and f - by

(1.4).

Then the function

z, ~{ f+(z)+f-(z*),

z E A ( a , • ) ,

f(z), z E

bA(a, Q),

is continuous on A(a, ~). In fact, on

A ( a , Q)

it coincides with the Poisson integral o f f .

We want to show t h a t a generalization of this holds for continuous functions on the annulus A with zero means on circles surrounding the origin.

390 3osip G l o b e v n i k

Suppose that f is a continuous function on A which satisfies (1.5) whenever

bA(a, o) c A

surrounds the origin. Recall that for each nonnegative integer n there are homogeneous polynomials P,, and Qn of degree n such that putting

z=re i~

into

~X3 2X2

1_ E Pn(z, 1 / 5 ) +

1 ~ Qn(L l/z)

Z Z

rt~O n = 0

we get the Fourier series of the function

ei~176

r l

<_r<_r2.

We now show that one can define a holomorphic function F + on ft § (A) by

(6.1)

F + ( z , w ) = l ~-~ Pn(z. 1/w). (z.w) Efl+(A),

W n = 0

where the series converges uniformly on compact sets in ft + (A) and a holomorphic function F - on f t - ( A ) by

(6.2) F - (z,

w) = 1 ~ Q,,(u,. 1/z).

(z, w) e f l - (A),

Z n = 0

where the series converges uniformly on compact sets in f~-(A).

Recall that for each t, 0 < t < l , the series (1/2) Y~=I

tJ+lPJ( z,

1/2) converges uniformly on A which, by a reasoning similar to the one in Section 5 implies that the function

~ c

(6.3) Ft+(z.w) = 1

-- ~-~,)+i,, ~ rjtz.

l / w )

IL' j = 0

is well defined, bounded and continuous on

f~*(A)tAbf~+(A)

and holomorphic on

~+(A) since the series (6.3) converges uniformly on

fF(A)Ubft+(A).

We have

F~-(z,

2~)=f~(z),

zEA.

Similarly, the function

F t- (z, w)= z E t~+lQJ

(w,

1/z)

j = O

is well defined, bounded and continuous on

ft-(A)Ubfl (A)

and holomorphic on f~- (A) since the series converges uniformly on ft- (A) Ubf~- (A). We have F t (z, 2) =

fi-(z), zEA.

Using the homogeneity of

Pj

we rewrite (6.3) to

1 pc

Ft'(z,

w) = ~ ~

Pj(tz, 1/(w/t))

j = O

A d e c o m p o s i t i o n o f f u n c t i o n s w i t h z e r o m e a n s o n c i r c l e s 3 9 1

and we see that (6.1) converges uniformly on Tt(f~+(A)), where Tt(z, w)=(tz, w/t).

Given a compact set KCf~+(A) there is a t. 0 < t < l . such that KcTt(Ft+(A)), so (6.1) converges uniformly on compact sets in f~*(A). We have

F ~ - ( z , w ) = r ( T , ( z , w ) ) , (z,u')ET[-I(~+(A)).

Since T, converges uniformly to the identity as t - + l it follows that (6.4) F+(z, w) = lim F?(z, w), (z. w) E Q'(A).

t - + l

where the convergence is uniform on compact sets in Q+(A). In the same way we see that

(6.5) F - ( z , w ) = l i m F { - ( z , w ) , (z. w) e f F ( A ) , where the convergence is uniform on compact sets in Q-(A).

One can verify that for each r, rl<<_r<r2, and for each t, 0 < t < l ,

ft(rei~176 OER.

where 7~ is the Poisson integral of the function ei~176 Now. 7~,.(tei~

f ( r e ~~ uniformly in 0 as t / ~ l and since f is uniformly continuous on A, the standard proof of the boundary continuity of the Poisson integral shows that the convergence is uniform also in r, ra ~ r ~ r 2 . Thus,

(6.6) ft--+ f uniformly on A as t / ~ 1.

7 . C o n t i n u o u s f u n c t i o n s w i t h z e r o m e a n s o n c i r c l e s

T h e o r e m 7.1. Let f be a continuous function on A which satisfies (1.5) for each bA(a, Q)c A which surrounds the origin. There are a holomorphic function F + on ~+(A) and a holomorphic function F - on Q (A) such that the function

{ F+(z,w)+F-(W. 5). (z,w) ef~+(A), (7.1) (z, w), >

f(z), (z, 5) C A.

has a bounded continuous extension to Q+ ( A )UbI~+ ( A ).

Thus, for each z E A we have

f(z) = lim [F+(~, ~) + F - (0, ~)].

(~,,)-~(z,~) ( ~,rI )CVI+ ( A )

392 Josip Globevnik

T h e o r e m 7.1 is the analogue of L e m m a 6.1 where

bA(a, 8)

is replaced by A and A(a, 8) is replaced by ~+(A). T h e function (7.1) is the analogue of the Poisson integral of f . It is the bounded continuous extension of

F(z, 2)=f(z)

to fl+(A)U

b~+(A)

which is pluriharmonic on fl+(A).

Proof of Theorem

7.1. For each t, 0 < t < l , define

r (z,w) EQ+(A)UbFt+(A),

where Ft + and F t- are as in Section 6. The properties of Ft + and F t imply that ~t is bounded and continuous on f~+

(A)Ub~+(A),

pluriharmonic on f~+ (A) and satisfies

qtt(z, 2)=f~(z)+f[(z)=ft(z), zEA.

It follows that for each t, s, 0 < t < l , 0 < s < l the function

(7.2)

(z,w), ) l ~ ( z , w ) - ~ s ( z , w ) l ,

restricted to Aa,~UbAa,e, attains its maximum on bAa,e whenever bA(a, Q)C A sur- rounds the origin. This is so since Aa,Q is biholomorphically equivalent to the punctured disc and since isolated singularities are removable for bounded harmonic functions. Thus, the function (7.2) attains its maximum on

~+(A)Ub~+(A)

on ,4.

By (6.6) the restrictions of functions kot to A converge uniformly as t / Z l which, by the preceding discussion implies that as t / Z l the functions ~ t converge uniformly on

~+(A)Ub~+(A)

to a bounded continuous function 9 which is pluriharmonic on f~+(A) and which satisfies

~P(z, 2)=f(z), zEA.

Now (6.4) and (6.5) imply that

tP(z,w)=F+(z,w)+F-(z,w), (z,w)Ef~+(A),

where F + and F - are given by (6.1) and (6.2). This completes the proof.

Acknowledgement.

This work was supported in part by a grant from the Min- istry of Education, Science and Sport of the Republic of Slovenia.

R e f e r e n c e s

[EK] EPSTEIN, C. L. and KLEINER, B., Spherical means in annular regions,

Comm. Pure Appl. Math.

46 (1993), 441-451.

[G1] GLOBEVNIK, J., Zero integrals on circles and characterizations of harmonic and analytic functions,

Trans. Amer. Math. Soc.

317 (1990), 313 330.

[G2] GLOBEVN1K, J., Holomorphic extensions from open families of circles,

Trans. Amer.

Math. Soe.

355 (2003), 1921-1931.

[M] MUSKHELISHVILI, N. I.,

Singular Integral Equations,

Noordhoff, Groningen, 1959.

A decomposition of functions with zero means on circles 393

[V] VOLCHKOV, V. V., Spherical means on Euclidean spaces. Ukrain. Mat. Zh. 50 (1998), 1310-1315 (Russian). English transl.: Ukrainian Math. J. 50 (1998), 1496-1503.

[Z] ZYGMUND, A., Trigonometric Series, Cambridge Univ. Press. 1959.

Received August 18, 2003 Josip Globevnik

Institute of Mathematics, Physics and Mechanics University of Ljubljana

Ljubljana Slovenia

email: josip.globevnik~fmf.uni-lj.si